# Hom between Brody hyperbolic varieties

If $$X$$ and $$Y$$ are smooth projective Brody hyperbolic varieties is $$\mathrm{Hom}(X, Y)$$ also Brody hyperbolic?

• Naive question: is there an example where this Hom is infinite? – Piotr Achinger Mar 2 at 13:31
• @PiotrAchinger maps from a point to a curve – user175135 Mar 2 at 13:33
• Remove constant maps then. – Piotr Achinger Mar 2 at 13:37
• @PiotrAchinger then take maps from $C$ to $C\times C$ – user175135 Mar 2 at 13:40
• Clearly I was being confused, incorrectly extrapolating from stuff we know (e.g. finite automorphism groups of varieties of general type, finite sets of conconstant maps to hyperbolic curves). The question now has a nice complete answer. That said, it seems that ${\rm Hom}(X,Y)$ could be understood more explicitly. For example, if a Brody hyperbolic surface $S$ admits infinitely many maps from some curve $C$, then I guess that $S$ must admit a generically finite rational map from $C\times D$ for some curve $D$. Decompose ${\rm Hom}(X,Y)$ by dimension of image, what can we say about the pieces? – Piotr Achinger Mar 2 at 19:29

## 1 Answer

I assume that for $$\operatorname{Hom}(X,Y)$$ you mean $$\operatorname{Hol}(X,Y)$$, that is the family of all holomorphic maps from $$X$$ to $$Y$$, endowed with its universal complex structure (which exists since your $$X$$ is compact).

As you probably know, for a compact complex space being Kobayashi hyperbolic is equivalent to being Brody hyperbolic.

Now, Theorem (6.4.1) in S. Kobayashi "Hyperbolic complex spaces" gives you the answer. Among other things it is stated there what follows.

Even if you assume $$X,Y$$ to be merely compact complex spaces (no need of any projectivity assumption), and only $$Y$$ to be hyperbolic, then $$\operatorname{Hol}(X,Y)$$ is compact and any of its connected component is compact hyperbolic (and hence Brody hyperbolic).

Aside comment (related to Piotr Achinger): if you look at the space of surjective holomorphic maps (or, more generally, dominant meromorphic maps), then under the same assumption as above, it is finite (Theorem (6.6.2) op. cit.).

• I meant the $\mathrm{Hom}$ functor which is representable by a scheme under the assumptions. But I think it coincides with what you wrote. – user175135 Mar 2 at 13:44
• Yes, I think the same. – diverietti Mar 2 at 13:50
• P.S. Since you are a new user and maybe you don't know exactly how it works here, may I say that once you get an answer which is satisfactory to you, then you should click to "accept" it. – diverietti Mar 2 at 13:58
• Absolutely. I just need some time to absorb it. – user175135 Mar 2 at 13:59
• I didn't want to be pushy of course! It was just in case as a new user you didn't know exactly how it works here! Take your time, and feel free not to "accept" my answer of course! :) – diverietti Mar 2 at 14:02